Posted by Anonymous on Saturday, July 27, 2013 at 6:55am.
the dimensions of a rectangle are such that its length is 5 in more than its width. If the length were doubled and the width were decreases by 2 in, the area would be increased 136 in^2,what are the length and width of the rectangle
- algebra - MathMate, Saturday, July 27, 2013 at 7:19am
Solve for w:
2(w-2)=w (w cannot be equal to -5)
- algebra - Reiny, Saturday, July 27, 2013 at 8:27am
original width -- x
original length -- x+5
area = x(x+5)
new width = x-2
new length = 2x+10
new area = (2x+10)(x-2)
(2x+10)(x-2) - x(x+5) = 136
2x^2 + 6x - 20 - x^2 - 5x - 136 = 0
x^2 + x - 156 = 0
(x-12)(x+13) = 0
x = 12 or x = -13
ignoring the negative,
width = 12
length = 17
original area = 12(17) = 204
new width = 10
new length = 34
new area = 10(34) = 340
increase in area = 340-204 = 136 , YEahh!
- algebra - MathMate, Saturday, July 27, 2013 at 4:23pm
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